Question:
Let $[t]$ denote the greatest integer $\leq t$. If for some
$\lambda \in \mathrm{R}-\{0,1\}, \lim _{\mathrm{x} \rightarrow 0}\left|\frac{1-\mathrm{x}+|\mathrm{x}|}{\lambda-\mathrm{x}+[\mathrm{x}]}\right|=\mathrm{L}$, then $\mathrm{L}$ is
equal to :
Correct Option: , 2
Solution:
$\mathrm{LHL}: \lim _{\mathrm{x} \rightarrow 0^{-}}\left|\frac{1-\mathrm{x}-\mathrm{x}}{\lambda-\mathrm{x}-1}\right|=\left|\frac{1}{\lambda-1}\right|$
$\mathrm{RHL}: \lim _{x \rightarrow 0^{+}}\left|\frac{1-x+x}{\lambda-x+1}\right|=\left|\frac{1}{\lambda}\right|$
For existence of limit
$\mathrm{LHL}=\mathrm{RHL}$
$\Rightarrow \frac{1}{|\lambda-1|}=\frac{1}{|\lambda|} \Rightarrow \lambda=\frac{1}{2}$
$\therefore \quad \mathrm{L}=\frac{1}{|\lambda|}=2$